0.000 000 000 000 031 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 031(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 000 031(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 031.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 031 × 2 = 0 + 0.000 000 000 000 062;
  • 2) 0.000 000 000 000 062 × 2 = 0 + 0.000 000 000 000 124;
  • 3) 0.000 000 000 000 124 × 2 = 0 + 0.000 000 000 000 248;
  • 4) 0.000 000 000 000 248 × 2 = 0 + 0.000 000 000 000 496;
  • 5) 0.000 000 000 000 496 × 2 = 0 + 0.000 000 000 000 992;
  • 6) 0.000 000 000 000 992 × 2 = 0 + 0.000 000 000 001 984;
  • 7) 0.000 000 000 001 984 × 2 = 0 + 0.000 000 000 003 968;
  • 8) 0.000 000 000 003 968 × 2 = 0 + 0.000 000 000 007 936;
  • 9) 0.000 000 000 007 936 × 2 = 0 + 0.000 000 000 015 872;
  • 10) 0.000 000 000 015 872 × 2 = 0 + 0.000 000 000 031 744;
  • 11) 0.000 000 000 031 744 × 2 = 0 + 0.000 000 000 063 488;
  • 12) 0.000 000 000 063 488 × 2 = 0 + 0.000 000 000 126 976;
  • 13) 0.000 000 000 126 976 × 2 = 0 + 0.000 000 000 253 952;
  • 14) 0.000 000 000 253 952 × 2 = 0 + 0.000 000 000 507 904;
  • 15) 0.000 000 000 507 904 × 2 = 0 + 0.000 000 001 015 808;
  • 16) 0.000 000 001 015 808 × 2 = 0 + 0.000 000 002 031 616;
  • 17) 0.000 000 002 031 616 × 2 = 0 + 0.000 000 004 063 232;
  • 18) 0.000 000 004 063 232 × 2 = 0 + 0.000 000 008 126 464;
  • 19) 0.000 000 008 126 464 × 2 = 0 + 0.000 000 016 252 928;
  • 20) 0.000 000 016 252 928 × 2 = 0 + 0.000 000 032 505 856;
  • 21) 0.000 000 032 505 856 × 2 = 0 + 0.000 000 065 011 712;
  • 22) 0.000 000 065 011 712 × 2 = 0 + 0.000 000 130 023 424;
  • 23) 0.000 000 130 023 424 × 2 = 0 + 0.000 000 260 046 848;
  • 24) 0.000 000 260 046 848 × 2 = 0 + 0.000 000 520 093 696;
  • 25) 0.000 000 520 093 696 × 2 = 0 + 0.000 001 040 187 392;
  • 26) 0.000 001 040 187 392 × 2 = 0 + 0.000 002 080 374 784;
  • 27) 0.000 002 080 374 784 × 2 = 0 + 0.000 004 160 749 568;
  • 28) 0.000 004 160 749 568 × 2 = 0 + 0.000 008 321 499 136;
  • 29) 0.000 008 321 499 136 × 2 = 0 + 0.000 016 642 998 272;
  • 30) 0.000 016 642 998 272 × 2 = 0 + 0.000 033 285 996 544;
  • 31) 0.000 033 285 996 544 × 2 = 0 + 0.000 066 571 993 088;
  • 32) 0.000 066 571 993 088 × 2 = 0 + 0.000 133 143 986 176;
  • 33) 0.000 133 143 986 176 × 2 = 0 + 0.000 266 287 972 352;
  • 34) 0.000 266 287 972 352 × 2 = 0 + 0.000 532 575 944 704;
  • 35) 0.000 532 575 944 704 × 2 = 0 + 0.001 065 151 889 408;
  • 36) 0.001 065 151 889 408 × 2 = 0 + 0.002 130 303 778 816;
  • 37) 0.002 130 303 778 816 × 2 = 0 + 0.004 260 607 557 632;
  • 38) 0.004 260 607 557 632 × 2 = 0 + 0.008 521 215 115 264;
  • 39) 0.008 521 215 115 264 × 2 = 0 + 0.017 042 430 230 528;
  • 40) 0.017 042 430 230 528 × 2 = 0 + 0.034 084 860 461 056;
  • 41) 0.034 084 860 461 056 × 2 = 0 + 0.068 169 720 922 112;
  • 42) 0.068 169 720 922 112 × 2 = 0 + 0.136 339 441 844 224;
  • 43) 0.136 339 441 844 224 × 2 = 0 + 0.272 678 883 688 448;
  • 44) 0.272 678 883 688 448 × 2 = 0 + 0.545 357 767 376 896;
  • 45) 0.545 357 767 376 896 × 2 = 1 + 0.090 715 534 753 792;
  • 46) 0.090 715 534 753 792 × 2 = 0 + 0.181 431 069 507 584;
  • 47) 0.181 431 069 507 584 × 2 = 0 + 0.362 862 139 015 168;
  • 48) 0.362 862 139 015 168 × 2 = 0 + 0.725 724 278 030 336;
  • 49) 0.725 724 278 030 336 × 2 = 1 + 0.451 448 556 060 672;
  • 50) 0.451 448 556 060 672 × 2 = 0 + 0.902 897 112 121 344;
  • 51) 0.902 897 112 121 344 × 2 = 1 + 0.805 794 224 242 688;
  • 52) 0.805 794 224 242 688 × 2 = 1 + 0.611 588 448 485 376;
  • 53) 0.611 588 448 485 376 × 2 = 1 + 0.223 176 896 970 752;
  • 54) 0.223 176 896 970 752 × 2 = 0 + 0.446 353 793 941 504;
  • 55) 0.446 353 793 941 504 × 2 = 0 + 0.892 707 587 883 008;
  • 56) 0.892 707 587 883 008 × 2 = 1 + 0.785 415 175 766 016;
  • 57) 0.785 415 175 766 016 × 2 = 1 + 0.570 830 351 532 032;
  • 58) 0.570 830 351 532 032 × 2 = 1 + 0.141 660 703 064 064;
  • 59) 0.141 660 703 064 064 × 2 = 0 + 0.283 321 406 128 128;
  • 60) 0.283 321 406 128 128 × 2 = 0 + 0.566 642 812 256 256;
  • 61) 0.566 642 812 256 256 × 2 = 1 + 0.133 285 624 512 512;
  • 62) 0.133 285 624 512 512 × 2 = 0 + 0.266 571 249 025 024;
  • 63) 0.266 571 249 025 024 × 2 = 0 + 0.533 142 498 050 048;
  • 64) 0.533 142 498 050 048 × 2 = 1 + 0.066 284 996 100 096;
  • 65) 0.066 284 996 100 096 × 2 = 0 + 0.132 569 992 200 192;
  • 66) 0.132 569 992 200 192 × 2 = 0 + 0.265 139 984 400 384;
  • 67) 0.265 139 984 400 384 × 2 = 0 + 0.530 279 968 800 768;
  • 68) 0.530 279 968 800 768 × 2 = 1 + 0.060 559 937 601 536;
  • 69) 0.060 559 937 601 536 × 2 = 0 + 0.121 119 875 203 072;
  • 70) 0.121 119 875 203 072 × 2 = 0 + 0.242 239 750 406 144;
  • 71) 0.242 239 750 406 144 × 2 = 0 + 0.484 479 500 812 288;
  • 72) 0.484 479 500 812 288 × 2 = 0 + 0.968 959 001 624 576;
  • 73) 0.968 959 001 624 576 × 2 = 1 + 0.937 918 003 249 152;
  • 74) 0.937 918 003 249 152 × 2 = 1 + 0.875 836 006 498 304;
  • 75) 0.875 836 006 498 304 × 2 = 1 + 0.751 672 012 996 608;
  • 76) 0.751 672 012 996 608 × 2 = 1 + 0.503 344 025 993 216;
  • 77) 0.503 344 025 993 216 × 2 = 1 + 0.006 688 051 986 432;
  • 78) 0.006 688 051 986 432 × 2 = 0 + 0.013 376 103 972 864;
  • 79) 0.013 376 103 972 864 × 2 = 0 + 0.026 752 207 945 728;
  • 80) 0.026 752 207 945 728 × 2 = 0 + 0.053 504 415 891 456;
  • 81) 0.053 504 415 891 456 × 2 = 0 + 0.107 008 831 782 912;
  • 82) 0.107 008 831 782 912 × 2 = 0 + 0.214 017 663 565 824;
  • 83) 0.214 017 663 565 824 × 2 = 0 + 0.428 035 327 131 648;
  • 84) 0.428 035 327 131 648 × 2 = 0 + 0.856 070 654 263 296;
  • 85) 0.856 070 654 263 296 × 2 = 1 + 0.712 141 308 526 592;
  • 86) 0.712 141 308 526 592 × 2 = 1 + 0.424 282 617 053 184;
  • 87) 0.424 282 617 053 184 × 2 = 0 + 0.848 565 234 106 368;
  • 88) 0.848 565 234 106 368 × 2 = 1 + 0.697 130 468 212 736;
  • 89) 0.697 130 468 212 736 × 2 = 1 + 0.394 260 936 425 472;
  • 90) 0.394 260 936 425 472 × 2 = 0 + 0.788 521 872 850 944;
  • 91) 0.788 521 872 850 944 × 2 = 1 + 0.577 043 745 701 888;
  • 92) 0.577 043 745 701 888 × 2 = 1 + 0.154 087 491 403 776;
  • 93) 0.154 087 491 403 776 × 2 = 0 + 0.308 174 982 807 552;
  • 94) 0.308 174 982 807 552 × 2 = 0 + 0.616 349 965 615 104;
  • 95) 0.616 349 965 615 104 × 2 = 1 + 0.232 699 931 230 208;
  • 96) 0.232 699 931 230 208 × 2 = 0 + 0.465 399 862 460 416;
  • 97) 0.465 399 862 460 416 × 2 = 0 + 0.930 799 724 920 832;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 031(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 1011 1001 1100 1001 0001 0000 1111 1000 0000 1101 1011 0010 0(2)

5. Positive number before normalization:

0.000 000 000 000 031(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 1011 1001 1100 1001 0001 0000 1111 1000 0000 1101 1011 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 45 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 031(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 1011 1001 1100 1001 0001 0000 1111 1000 0000 1101 1011 0010 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 1011 1001 1100 1001 0001 0000 1111 1000 0000 1101 1011 0010 0(2) × 20 =

1.0001 0111 0011 1001 0010 0010 0001 1111 0000 0001 1011 0110 0100(2) × 2-45


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -45

Mantissa (not normalized):
1.0001 0111 0011 1001 0010 0010 0001 1111 0000 0001 1011 0110 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-45 + 2(11-1) - 1 =

(-45 + 1 023)(10) =

978(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 978 ÷ 2 = 489 + 0;
  • 489 ÷ 2 = 244 + 1;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

978(10) =

011 1101 0010(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0111 0011 1001 0010 0010 0001 1111 0000 0001 1011 0110 0100 =

0001 0111 0011 1001 0010 0010 0001 1111 0000 0001 1011 0110 0100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0010

Mantissa (52 bits) =
0001 0111 0011 1001 0010 0010 0001 1111 0000 0001 1011 0110 0100


Decimal number 0.000 000 000 000 031 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0010 - 0001 0111 0011 1001 0010 0010 0001 1111 0000 0001 1011 0110 0100

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100